Answer to Question #320921 in Statistics and Probability for Deshna

Question

A population of 1,000 students has an average weekly allowance of μ = 350 Php and standard deviation of σ = 56.13 Php. What is the probability that a random sample of size n = 30 will have an average weekly allowance between 335 and 360 Php?

Answer 1

Let X be a random variable representing the sample mean of size 30, then X ~ $N(350, ({\frac {56.13^2} {30}})^2)$

$P(335<X<360)=P(335<N(350, ({\frac {56.13^2} {30}})^2)<360)=P(335<350+10.25Z<360)=P(-1.463<Z<0.976)=P(Z<0.976)-P(Z<-1.463)=0.83547-0.07173=0.76374$